Page 50 - 35Linear Algebra

50 Systems of Linear Equations

has no solutions. If you remove one of the rows of this matrix, does
the new matrix have any solutions? In general, can row equivalence be
affected by removing rows? Explain why or why not.

5. Explain why the linear system has no solutions:

 
1 0 3 1
0 1 2 4
 
0 0 0 6

For which values of k does the system below have a solution?

x − 3y = 6
x + 3z = − 3
2x + ky + (3 − k)z = 1

Hint

6. Show that the RREF of a matrix is unique. (Hint: Consider what
happens if the same augmented matrix had two different RREFs. Try
to see what happens if you removed columns from these two RREF
augmented matrices.)

7. Another method for solving linear systems is to use row operations to
bring the augmented matrix to Row Echelon Form (REF as opposed to
RREF). In REF, the pivots are not necessarily set to one, and we only
require that all entries left of the pivots are zero, not necessarily entries
above a pivot. Provide a counterexample to show that row echelon form
is not unique.
Once a system is in row echelon form, it can be solved by “back substi-
tution.” Write the following row echelon matrix as a system of equa-
tions, then solve the system using back-substitution.

 
2 3 1 6
0 1 1 2
 
0 0 3 3

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